DIAMETER PROBLEMS FOR UNIVALENT FUNCTIONS WITH QUASICONFORMAL EXTENSION

This paper utilizes the method of extremal length to study several diameter problems for functions conformal outside of a disc centered at the origin, with a standard normalization, which possess a quasiconformal extension to a ring subdomain of this disc. Known results on the diameter of a complementary component of the image domain of a univalent function are extended. Applications to the transfinite diameters of families of non-overlapping functions and an extension of the Koebe one-quarter theorem are included.

For an arbitrary domain/ that contains infinity, let X:() consist of all functions y such that f is univalent and regular in apart from a simple pole at infinity where it has the normalized expansion .f(z)z + bo + blz -+ For a simple closed Jordan curvecontained in 9, the class X; (:o) consists of all functions f such that f restricted to n Ext(-y) is in E( n Ext(.)) j' rcstricted K,'y to :O Int(,) is K-qc., and y is univalent in :O-{'r}.Note that f need not be continuous (or even defined) on "r, so one could consider f as a pair of "non-overlapping" functions defined on the pair of domains f Ext(,) and/ Int (7).If the d (for discontinuous) is dropped from the symbol for the above class of functions, then we require continuity on 7. It is clear that EK,(:D) C EK,(D).As a special case, if "r is the circle Cn {[z[ R}, then let X;.,n represent the family :EK,c(-CI(D)).
Modules of a curve family in a domain will be used in this study, and will be denoted by re(V) when the curve family has been specified.For reference to this item, see Jenkins [2].The following canonical mappings are also used.For a domain :o of finite connectivity which contains the origin, let O1 denote the circular slit disc mapping which fixes the origin and with O](0) > 0, and for a domain of finite connectivity which has at least two boundary continua, let o2 represent "the" circular slit ring mapping, which is unique up to rotations.We also use the radial slit disc mapping which fixes the origin and has positive first derivative at the origin.The existence of these mappings is well-known, and our reference will be Jenkins [2], theorems 5.4, 5.6, and 5.5, respectively, pages 74-75.
3. DIAMETER THEOREMS.THEOREM 1.Let :O be a domain on the w-sphere which contains the point at infinity with boundaxy components E1 C/(D), E:,...,E,.,,where the E, for j 2 n, are slits on concentric circles centered at the origin.Let I be univalent in 9, K-qc. in D1 c {1 < Iw] </}, regular in :O0 :O C {Iw] >/} apart from a simple pole at infinity where it has the expansion f(to) OtO + a0 + (1110 -1 + The function f need not be continuous on Ca {Iwl p}, but the complementary component of the image domain of I corresponding to Ex must contain CI(D).Then [a[ >_ pxm-x, with equality if d only if I is equal to e*,p/x-w for Iwl > P, ei'wc+(u) -(K) for < ]w] < p, where , R.
PROOF.First we note that, since our goal is to minimize ]al, we may restrict our mappings I to those su that El C/(D) and where there are no "gaps" betwn D and D, where image domains and complentary components are denoted by prim.This statement arises from application of the emann mapping threm combined with Swarz's inequality.
Let (w) log(w), where the branch chosen is that which takes imaginary values betwn 0 d 2. First we cut off a neighborhood of infinity.Let (R) {]w] < R}, where R is osen large enough so that the boundary components of 9 are contained in/)(R).Next one transfers to the logarithmic plane via the mapping (.Let A(R)= ((/)(R)), AI ((190, and The function I induces a mapping w F() on the variable ( as follows: start with a value of ( E A(R), take the corresponding value of w, perform the mapping l(w) w', then map this point by the same branch of .Ac alculation reveals that, on the segment ai {(, ) 0 _< < 2, log(R)}, F() C + os(I-I) + o(,-).
We estimate the module of A'(R) for the class of curves which separate the boundary component El from the image of as under F.
For a lower bound, we employ a slight extension of GrStzsch's lemma (see Jenkins[2], theorem 2.6, page 22).In our case, the domains A'(R), A], and A(R) are not doubly-connected due to the additional boundary components.To adapt GrStzsch's lemma to the present situation, instead of employing the canonical mapping of a doubly-connected domain onto a circular ring, as in the proof of GrStzsch's lemma, one must employ the circular slit ring mapping @2.With this modification, the rest of the proof of GrStzsch's lemma carries through.After this, we apply the quasi-invariance of the module under a K-qc.mapping and conformal invariance of the module to show that To estimate the module from above, the metric (2) -1 [d(l is clearly admissible in the module problem, thus by employing the expansion for F on aa we have m(A'(R)) _< (2r) -2 //dA '() _< (-)-[og(I-I) + log(a) + O(-)].
Combining the upper and lower estimates and letting R tend to infinity yields the inequality of the theorem.
The equality statement is obtained by first assuming equality holds throughout the proof and then employing the equality statement in Gr6tzsch's (extended) lemma.For w )0, the uniqueness of the circular slit mapping for domains of finite connectivity (theorem 5.4, page 74, Jenkins [2]) is used, and for w 6 a, one proceeds as in Kfinzi [5], page 100, Satz 3a.
We remark that the proof of the inequality in theorem I only depends upon the modules of and A0(R).Thus the inequality remains valid for domains of infinite connectivity, provided the set of boundary slits do not affect the modules of A and A0(R), i.e., the set of boundary slits form a minimal slit set (for refe:ence, see Jenkins [2], page 81).The price one pays for this extension is a sacrifice of uniqueness of the extremal mapping, since the circular slit mapping need not be unique for domains of infinite connectivity.
We mention another extension of theorem 1.Its proof is an obvious modification of the proof of theorem 1.For theorem 2, q will always represent any integer between and N, inclusive, for N any fixed integer.Let R0 < R < ..-< R, denote Cn {]w] R}, and let where z) is as in theorem 1. THEOREM 2. Let z) be a domain conditioned as in theorem 1.Let I be as in theorem 1, but now assume I is K-qc. in Z), I not necessarily continuous on C. Then I-I >-(,/,_)v, with equality if and only if f(w) A,,.,,,/,(w) where A,,, ,,+,(w) e'*+'R H=, (R/R-z) '/r w else, where z,... ,+ E R.
The next corollary is proven by forming a Riemann sum and applying the above theorem.For an example of this, see McLeavey  [6], and for another method of proof, see Kunzi [5], page 27.
COROLLARY.Let v be a domain conditioned in threm 1.Let I be in threm 1, but now sume that I is K(Iwl)-qc.in V {1 < Iwl < p} , where for simplicity, we sume that K(Iwl) is a continuous function.Then lal pexp [, K(r) with equality if and only if, for w r0e*, , e R, exp Jz K(r) eie i' for < r0 < p, l() L,g, for ro > p.
will be clear eorresponin results coul be proven usin rem 2 in place of heorem ].
Then there exists a mapping h e Z(-T') such that the complementary component of the image domain of h is a horizontal segment of length 4a, for some > 0, with The proof of this lemma follows easily from the Riemann mapping theorem and the classical maximum diameter theorem for univalent functions.Since h n eK is a competitor in our problem, to maximize diameter it suffices to assume that T' is a horizontal segment of length 4a, which we temporarily assume lies on the real axis and is centered at the origin.The goal now becomes to maximize a, which is accomplished by applying theorem to the function g-omllaonomqogoAo,o(w), where the branch of g-is that which maps onto the exterior of the closed unit disc, to yield a < Q.
The uniqueness statement is proven by removing our temporary assumption on T' and allowing rotations in 01 and %, and then renormalizing by rotating the final image domain.
When O; and o; are known, one can explicitly calculate the maximum diameter and all extrcmal functions, as in the following corollary.COROLLARY 1.Let 1'" e :E,R.The diameter e of the complementary component of the image domain of f" satisfies e < 4R1-1/K, with equality if and only if y" equals z + (R2)-lmz -1 for Izl >_ /E(z) and the rotations and translations e-iys(esz) + c, where c is an arbitrary constant.
A standard manipulation of corollary provides an extension of Koebe's one-quarter theorem.
COROLLARY 2. Let I" be univalent in D, regular in {Izl < r} with a continuous K-qc.extension to the rest of D, with an expansion in a neighborhood of the origin being y*(z) z + a2z2+.... Also assume that I" does not take the value infinity in D. Then the values of J" cover the disc {Iwl < 4-Irl-a/K}.If an omitted value w0 of y" is such that Iw01-4-1r1-I/K, then y" equals z/ (1 + z2e-=rO/x-l)) for Izl fS(z) (-z i,i)x/,-a,/(1 + -'is I,I '('/K-a) ,=) for r < Next an application of corollary 1 is proven which extends corollary 7.3, page 124, in Jenkins [2].It will be clear that one can also extend corollaries 7.1 and 7.4 in Jenkins [2].First some labels are given.Let 0 < r < < r < oo, let K, > 1, for 1, 2, let Let Sx consist of MI functions ft such that fx is K-qc. in Rt, with a not necsily continuous nformM extension to D, with expsion for z D1 ing ft(z) az + az + ..., d 1x is non- overlapping, ((}) f((D1})= $.Let E consist of MI functions f su that is K-qc. in R, with a not necessarily continuous conformal extension to -CI(D), with expansion for z e being y(z) bz + bo + bz -+..., and ({}) y({V}) $.PROOF.The proof follows the same scheme as before.One forms the alleged extremal functions, then composes them with functions such that the compositions are competitors in the extremal problem.The univalent case (Jenkins [2], corollary 7.3) combined with the Riemann mapping theorem allows one to restrict competition to only those functions whose image domains are geometrically nice.An application of corollary then proves the theorem.
We conclude this paper with corresponding results for the minimum diameter theorem.For this we first achieve a result analogous to theorem 1. THEOREM 4. Let /9 be a domain on the w-sphere which contains infinity, with boundary components E CI(D), E E, where the E, for j 2,...,n, consist of slits on rays emanating from the origin, and which are contained in {1 < Iwl < p}.Let f be continuous and univalent in/9 with continuous first partials in/9 =/ t3 {1 < Iwl < p}, regular in/90 79 t3 {Iwl > p} apart from a simple pole at infinity where it has the expansion (3.1).Assume the complementary component of the image domain of I which corresponds to Ex is the closed unit disc.Let ((w) log(w) be as in theorem 1, let A ((/ga), and let F(() be the mapping induced by as in theorem 1.Finally, for A1, assume the induced map F satisfies 0 + ("- (3.) Then lal < 1, with equality if and only if I is a rotation of 9.
PROOF.As in theorem 1 we map to the logarithmic plane.We will need uniform convergence of the expansion of F, so let > 0, let A, be the image of {1 < Iwl < '} n under , and let A', be the image of A, under the induced mapping F. For e A0 (Z0), the following expansion is valid for F: F() ff + los(c) + powers of -Using this expansion, an upper bound for the area of A', is given by .4,,:_< / [r(log(p) + + i)] ao where As denotes the area of a set S.
For a lower bound, first we must restrict our functions to those whose image domains are geometrically nice.LEMMA 2. Assume the complementary components of the image domain of f corresponding to /2 /, do not all consist of slits on rays emanating from the origin.Then there exists a function h, regular and univalent in f(9) apart from a simple pole at infinity where h(W)= flW+bo+baW -1 +..-, such that the complementary component of the image domain of h corresponding to the closed unit disc is the closed unit disc, the other complementary components consist of slits on rays emanating from the origin, and furthermore, Itl > 1.
The proof of this lemma is identical to the proof of lemma 1 except now the existence of h is a well-known example of a canonical conformal mapping (see, for example, theorem 5.5, Jenkins [2], page 74).It is also well-known that I#l > 1, strict inequality due to the assumption on the image domain of f (see theorem 5.2, Jenkins [2], page 73).
If we let g" hof, it is easy to see that the mapping induced by g" satisfies inequality (3.2).Thus g is a competitor of / with a geometrically nice image domain such that [g"(cx)l > lY()I.Therefore it suffices to assume that the image domain of f has complementary components consisting of the closed unit disc along with slits on rays emanating from the origin.
The lower bound for the area of A' is now derived.First integrate both sides of (3.2) over A. Apply Schwarz's inequality and then Fubini's theorem to each of the integrals on the right side of (3.2).After this, employ an obvious geometric inequality, which is where the lemma on the boundary components of the image domain of f comes into play.The details are as follows.
Similarly, using vertical segmen instead of horizontM segments, we get Inserting (a.a) aria (a.4) into the integrated inequality obtained from (a.2) yields ]] l e .+ ( + /( + ') Combining the upper end lower estimetes yields the inequality of the threm.The prf of the equality stetement is streightforwerd.COROLLARY.Let I" " The dieter 0 f the complement component f I" satisfies o R-, with equality i[ and only if I" equals I(.) for . ,R-g*(gl -(l for < I1 < R, end the trslations I()+ e, where e i en arbitrary PROOF.Let I be bitr funeti u& that I" I -, ] " ith expsin in {11 R} being I() R-+,+-+ nthe gue f le 1 i nded rriet our functions to those with gmegricy ni imege domn. he pmf i he e lee 1, except now one mbine he emn mpping herem wih the elie nimum diameter theorem for unilent functions.
LEMMA .Assume the complemeny component ' of the imege domin f responds C() is not e circle.Then there exists h (-') such thet the cmplemt mponent of the image domein of is e circle of diameter e. Morver, if 0" i the dieter then 0 < By lem a, since he goal i minie dieter, end ince h I" is competitor, ume thet I" h compleme component e circle centered e e poin e, f miniing becomes the o.A eMcultion verifie th inequality (2) f threm 4 hld [r the induced mepping F of I. Thus n eppliction of threm 4 to the function / T_ I proves the corollery.